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hai...Plz help me.......... I am creating my application using MDI form...I want to go to another form when a button clicks and the new form should be displayed in that mdi form..

Kirtan Patel · Answer

then Write Code in Agent Form's Load Event like below private void Agent_Load( object sender, EventArgs e) { //Create Instance of Another Form Say Form 2 Form2 f2 = new Form2 (); //Set MDI parent as Form1 ( Here Form1 is Main MDI Form) f2.MdiParent = Application .OpenForms[ "Form1" ]; //Show the Form f2.Show(); }

Soumya Surendran · Answer

Thanks a lot.............

Soumya Surendran · Answer

thank you for your reply.....I have opened the form1 using the following code agent ag = new agent (); ag.MdiParent = this ; menuStrip1.Enabled = false ; ag.Show(); from this form "agent" i want to go to form2 which should be displayed as the child of mdi form.....

Kirtan Patel · Answer

Hi , you can Do it this way :) private void button1_Click( object sender, EventArgs e) { //Create Instance of Another Form Say Form 2 Form2 f2 = new Form2 (); //Set MDI parent as Current Form f2.MdiParent = this ; //Show the Form f2.Show(); }

Niradhip Chakraborty · Answer

say you want to open childform(frmUsers ) from MDI from. private void mnuUsers_Click(object sender, System.EventArgs e) { foreach (Form frmuser in this.MdiChildren) { if (frmuser is frmUsers) { frmuser.Show(); frmuser.Focus(); return; } } frmUsers frmuser1 = new frmUsers(); this.AddOwnedForm(frmuser1); frmuser1.MdiParent = this; frmuser1.Show(); }

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